Prime avoidance theorem: Let $A$ be a ring (commutative with unity) and $p_1,...,p_n\subset A$ prime ideals. Let $a\subset A$ be an ideal such that $a\subset (p_1\cup p_2\cup\cdot\cdot\cdot\cup p_n)$, then $a\subset p_k$ for some $1\leq k\leq n$.

Now, I have no problem in proving this theorem. I want to illustrate with an example, the importance of the prime condition in the theorem. That is, I want to show that there exist ideals $a_1,...,a_n\subset A$ such that $a\subset (a_1\cup a_2\cup\cdot\cdot\cdot\cup a_n)$, but $a\not \subset a_k$ for all $1\leq k\leq n$. Some of the things that I noticed is that, we cannot find the example in a principal ideal domain nor can we get the example if we take $n=2$. I wasn't able to make much progress beyond this.

Thank you in advance!


2 Answers 2


Let $A=\mathbb {Z}[X,Y], a=(2,X,Y), a_{1}=(2,X^2,Y), a_{2}=(2,X,Y^2), a_{3}=(2,X+Y,X^2,Y^2,XY)$

Then $a$ is contained in the union of $a_1,a_2,a_3$ (you can check it by computing in $A/(2,X^2,Y^2,XY)$ , which is a ring with $4$ elements) but not contained in any $a_i$.

Source of the counter example: https://fr.wikipedia.org/wiki/Lemme_d%27%C3%A9vitement_des_id%C3%A9aux_premiers


Well, here is a rather standard example.

Let $V$ be a $m$-dimensional vector space over a finite field $k$, $2\leq m<\infty$. Put on $V$ a ring (without identity) structure where all multiplications are 0. Unitize it to give $A=k\oplus V$ a commutative ring (with 1). Let $J_1,J_2,\dots,J_n$ be the proper subspaces of $V$, so are non-prime ideals of $A$. Then the (maximal) ideal $V$ is not contained in any $J_i$ but is contained in their union.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .